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exercises.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Exercises}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Algebra}
\label{section-algebra}
\noindent
This first section just contains some assorted questions.
\begin{exercise}
\label{exercise-isomorphism-localizations}
Let $A$ be a ring, and ${\mathfrak m}$ a maximal ideal. In $A[X]$
let $\tilde {\mathfrak m}_1 = ({\mathfrak m}, X)$ and
$\tilde {\mathfrak m}_2 = ({\mathfrak m}, X-1)$. Show
that
$$
A[X]_{\tilde {\mathfrak m}_1} \cong A[X]_{\tilde {\mathfrak m}_2}.
$$
\end{exercise}
\begin{exercise}
\label{exercise-coherent}
Find an example of a non Noetherian ring $R$ such that every
finitely generated ideal of $R$ is finitely presented as an $R$-module.
(A ring is said to be {\it coherent} if the last property holds.)
\end{exercise}
\begin{exercise}
\label{exercise-flat-ideals-pid}
Suppose that $(A, {\mathfrak m}, k)$ is a Noetherian local ring. For any
finite $A$-module $M$ define $r(M)$ to be the minimum number
of generators of $M$ as an $A$-module. This number equals
$\dim_k M/{\mathfrak m} M = \dim_k M \otimes_A k$ by NAK.
\begin{enumerate}
\item Show that $r(M \otimes_A N) = r(M)r(N)$.
\item Let $I\subset A $ be an ideal with $r(I)>1$. Show that
$r(I^2) < r(I)^2$.
\item Conclude that if every ideal in $A$ is a flat module, then
$A$ is a PID (or a field).
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-not-isomorphic}
Let $k$ be a field. Show that the following pairs of
$k$-algebras are not isomorphic:
\begin{enumerate}
\item $k[x_1, \ldots, x_n]$ and $k[x_1, \ldots, x_{n + 1}]$ for any
$n\geq 1$.
\item $k[a, b, c, d, e, f]/(ab + cd + ef)$ and $k[x_1, \ldots, x_n]$
for $n = 5$.
\item $k[a, b, c, d, e, f]/(ab + cd + ef)$ and $k[x_1, \ldots, x_n]$
for $n = 6$.
\end{enumerate}
\end{exercise}
\begin{remark}
\label{remark-simple-geometric}
Of course the idea of this exercise is to find
a simple argument in each case rather than applying a ``big'' theorem.
Nonetheless it is good to be guided by general principles.
\end{remark}
\begin{exercise}
\label{exercise-silly}
Algebra. (Silly and should be easy.)
\begin{enumerate}
\item Give an example of a ring $A$ and a nonsplit
short exact sequence of $A$-modules
$$
0 \to M_1 \to M_2 \to M_3 \to 0.
$$
\item Give an example of a nonsplit sequence of $A$-modules
as above and a faithfully flat $A \to B$ such that
$$
0 \to M_1\otimes_A B \to M_2\otimes_A B \to M_3\otimes_A B \to 0.
$$
is split as a sequence of $B$-modules.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-field-kummer}
Suppose that $k$ is a field having a primitive $n$th root
of unity $\zeta$. This means that $\zeta^n = 1$, but $\zeta^m\not = 1$ for
$0<m<n$.
\begin{enumerate}
\item Show that the characteristic of $k$ is prime to $n$.
\item Suppose that $a \in k$ is an element of $k$ which is not
an $d$th power in $k$ for any divisor $d$ of $n$, $in\geq d>1$. Show that
$k[x]/(x^n-a)$ is a field. (Hint: Consider a splitting field for
$x^n-a$ and use Galois theory.)
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-valuation}
Let $\nu : k[x]\setminus \{0\} \to {\mathbf Z}$ be a map
with the following properties: $\nu(fg) = \nu(f) + \nu(g)$ whenever
$f$, $g$ not zero, and $\nu(f + g) \geq min(\nu(f), \nu(g))$ whenever
$f$, $g$, $f + g$ are not zero, and $\nu(c) = 0$ for all $c\in k^*$.
\begin{enumerate}
\item Show that if $f$, $g$, and $f + g$ are nonzero and
$\nu(f) \not= \nu(g)$ then we have equality $\nu(f + g) = min(\nu(f), \nu(g))$.
\item Show that if $f = \sum a_i x^i$, $f\not = 0$, then
$\nu(f) \geq min(\{i\nu(x)\}_{a_i\not = 0})$. When does equality hold?
\item Show that if $\nu$ attains a negative value then
$\nu(f) = -n \deg(f)$ for some $n\in {\mathbf N}$.
\item Suppose $\nu(x) \geq 0$. Show that
$\{f \mid f = 0, \ or\ \nu(f) > 0\}$ is a prime ideal of $k[x]$.
\item Describe all possible $\nu$.
\end{enumerate}
\end{exercise}
\noindent
Let $A$ be a ring. An {\it idempotent} is an element $e \in A$
such that $e^2 = e$. The elements $1$ and $0$ are always idempotent.
A {\it nontrivial idempotent} is an idempotent which is not
equal to zero. Two idempotents $e, e' \in A$ are called {\it orthogonal}
if $ee' = 0$.
\begin{exercise}
\label{exercise-product}
Let $A$ be a ring. Show that $A$ is a product of two nonzero rings if
and only if $A$ has a nontrivial idempotent.
\end{exercise}
\begin{exercise}
\label{exercise-lift-idempotents}
Let $A$ be a ring and let $I \subset A$ be a locally nilpotent ideal.
Show that the map $A \to A/I$ induces a bijection on idempotents.
(Hint: It may be easier to prove this when $I$ is nilpotent. Do this first.
Then use ``absolute Noetherian reduction'' to reduce to the nilpotent case.)
\end{exercise}
\section{Colimits}
\label{section-colimits}
\begin{definition}
\label{definition-directed-poset}
A {\it directed partially ordered set} is a nonempty set $I$ endowed
with a partial ordering $\leq$ such that given any pair $i, j \in I$
there exists a $k \in I$ such that $i \leq k$ and $j \leq k$. A
{\it system of rings} over $I$ is given by a ring $A_i$ for each $i \in I$
and a map of rings $\varphi_{ij} : A_i \to A_j$ whenever $i \leq j$ such that
the composition $A_i \to A_j \to A_k$ is equal to
$A_i \to A_k$ whenever $i \leq j \leq k$.
\end{definition}
\noindent
One similarly defines systems of groups, modules over a fixed ring,
vector spaces over a field, etc.
\begin{exercise}
\label{exercise-directed-colimit}
Let $I$ be a directed partially ordered set and let
$(A_i, \varphi_{ij})$ be a system of rings over $I$.
Show that there exists a ring $A$ and maps $\varphi_i : A_i \to A$
such that $\varphi_j \circ \varphi_{ij} = \varphi_i$ for all $i \leq j$
with the following universal property: Given any ring $B$
and maps $\psi_i : A_i \to B$ such that
$\psi_j \circ \varphi_{ij} = \psi_i$ for all $i \leq j$, then
there exists a unique ring map $\psi : A \to B$ such that
$\psi_i = \psi \circ \varphi_i$.
\end{exercise}
\begin{definition}
\label{definition-colimit}
The ring $A$ constructed in Exercise \ref{exercise-directed-colimit}
is called the {\it colimit} of the system. Notation $\colim A_i$.
\end{definition}
\begin{exercise}
\label{exercise-prime-in-colimit}
Let $(I, \geq)$ be a directed partially ordered set and let
$(A_i, \varphi_{ij})$ be a system of rings over $I$ with colimit
$A$. Prove that there is a bijection
$$
\Spec(A) = \{(\mathfrak p_i)_{i \in I} \mid\
\mathfrak p_i \subset A_i \text{ and }
\mathfrak p_i = \varphi_{ij}^{-1}(\mathfrak p_j)\ \forall i \leq j\}
\subset \prod\nolimits_{i \in I} \Spec(A_i)
$$
The set on the right hand side is the limit of the sets
$\Spec(A_i)$. Notation $\lim \Spec(A_i)$.
\end{exercise}
\begin{exercise}
\label{exercise-colimit-surjective}
Let $(I, \geq)$ be a directed partially ordered set and let
$(A_i, \varphi_{ij})$ be a system of rings over $I$ with colimit
$A$. Suppose that $\Spec(A_j) \to \Spec(A_i)$ is
surjective for all $i \leq j$. Show that
$\Spec(A) \to \Spec(A_i)$ is surjective for all $i$.
(Hint: You can try to use Tychonoff, but there is also a basically trivial
direct algebraic proof based on
Algebra, Lemma \ref{algebra-lemma-in-image}.)
\end{exercise}
\begin{exercise}
\label{exercise-integral-colimit-finite}
Let $A \subset B$ be an integral ring extension. Prove that
$\Spec(B) \to \Spec(A)$ is surjective.
Use the exercises above, the fact that this holds for a finite ring
extension (proved in the lectures), and by proving that
$B = \colim B_i$ is a directed colimit of finite extensions $A \subset B_i$.
\end{exercise}
\begin{exercise}
\label{exercise-colimit-tensor}
Let $(I, \geq)$ be a partially ordered set which is directed.
Let $A$ be a ring and let $(N_i, \varphi_{i, i'})$ be a directed system of
$A$-modules indexed by $I$. Suppose that $M$ is another $A$-module. Prove
that
$$
\colim_{i\in I} M \otimes_A N_i\cong
M \otimes_A \Big( \colim_{i\in I} N_i\Big).
$$
\end{exercise}
\begin{definition}
\label{definition-finite-presentation}
A module $M$ over $R$ is said to be of {\it finite presentation} over
$R$ if it is isomorphic to the cokernel of a map of finite free modules
$ R^{\oplus n} \to R^{\oplus m}$.
\end{definition}
\begin{exercise}
\label{exercise-colimit-modules}
Prove that any module over any ring is
\begin{enumerate}
\item the colimit of its finitely generated submodules, and
\item in some way a colimit of finitely presented modules.
\end{enumerate}
\end{exercise}
\section{Additive and abelian categories}
\label{section-additive}
\begin{exercise}
\label{exercise-filtered-vector-spaces}
Let $k$ be a field. Let $\mathcal{C}$ be the category of filtered vector
spaces over $k$, see
Homology, Definition \ref{homology-definition-filtered}
for the definition of a filtered object of any category.
\begin{enumerate}
\item Show that this is an additive category (explain carefuly what the
direct sum of two objects is).
\item Let $f : (V, F) \to (W, F)$ be a morphism of $\mathcal{C}$.
Show that $f$ has a kernel and cokernel (explain precisely
what the kernel and cokernel of $f$ are).
\item Give an example of a map of $\mathcal{C}$ such that
the canonical map $\text{Coim}(f) \to \text{Im}(f)$ is not an isomorphism.\
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-torsion-free}
Let $R$ be a Noetherian domain. Let $\mathcal{C}$ be the category of
finitely generated torsion free $R$-modules.
\begin{enumerate}
\item Show that this is an additive category.
\item Let $f : N \to M$ be a morphism of $\mathcal{C}$.
Show that $f$ has a kernel and cokernel (make sure you define precisely
what the kernel and cokernel of $f$ are).
\item Give an example of a Noetherian domain $R$ and a map of
$\mathcal{C}$ such that the canonical map $\text{Coim}(f) \to \text{Im}(f)$
is not an isomorphism.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-other}
Give an example of a category which is additive and has kernels
and cokernels but which is not as in
Exercises \ref{exercise-filtered-vector-spaces} and
\ref{exercise-torsion-free}.
\end{exercise}
\section{Flat ring maps}
\label{section-flat}
\begin{exercise}
\label{exercise-localization-flat}
Let $S$ be a multiplicative subset of the ring $A$.
\begin{enumerate}
\item For an $A$-module $M$ show that $S^{-1}M = S^{-1}A \otimes_A M$.
\item Show that $S^{-1}A$ is flat over $A$.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-examples-not-flat}
Find an injection $M_1 \to M_2$ of $A$-modules such that
$M_1\otimes N \to M_2 \otimes N$ is not injective in the following
cases:
\begin{enumerate}
\item $A = k[x, y]$ and $N = (x, y) \subset A$. (Here and below $k$ is a field.)
\item $A = k[x, y]$ and $N = A/(x, y)$.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-flat-not-projective}
Give an example of a ring $A$ and a finite $A$-module $M$
which is a flat but not a projective $A$-module.
\end{exercise}
\begin{remark}
\label{remark-flat-not-projective}
If $M$ is of finite presentation and flat over $A$,
then $M$ is projective over $A$. Thus your example will have to
involve a ring $A$ which is not Noetherian. I know of an example
where $A$ is the ring of ${\mathcal C}^\infty$-functions on ${\mathbf R}$.
\end{remark}
\begin{exercise}
\label{exercise-flat-not-free-dvr}
Find a flat but not free module over ${\mathbf Z}_{(2)}$.
\end{exercise}
\begin{exercise}
\label{exercise-flat-deformations}
Flat deformations.
\begin{enumerate}
\item Suppose that $k$ is a field and $k[\epsilon]$ is the ring of
dual numbers $k[\epsilon] = k[x]/(x^2)$ and $\epsilon = \bar x$. Show that for
any $k$-algebra $A$ there is a flat $k[\epsilon]$-algebra $B$ such that
$A$ is isomorphic to $B/\epsilon B$.
\item Suppose that $k = {\mathbf F}_p = {\mathbf Z}/p{\mathbf Z}$ and
$$
A = k[x_1, x_2, x_3, x_4, x_5, x_6]/
(x_1^p, x_2^p, x_3^p, x_4^p, x_5^p, x_6^p).
$$
Show that there exists a flat ${\mathbf Z}/p^2{\mathbf Z}$-algebra $B$ such
that $B/pB$ is isomorphic to $A$. (So here $p$ plays the role of $\epsilon$.)
\item Now let $p = 2$ and consider the same question for
$k = {\mathbf F}_2 = {\mathbf Z}/2{\mathbf Z}$ and
$$
A = k[x_1, x_2, x_3, x_4, x_5, x_6]/
(x_1^2, x_2^2, x_3^2, x_4^2, x_5^2, x_6^2, x_1x_2 + x_3x_4 + x_5x_6).
$$
However, in this case show that there does {\it not} exist a flat
${\mathbf Z}/4{\mathbf Z}$-algebra $B$ such that $B/2B$ is isomorphic to
$A$. (Find the trick! The same example works in arbitrary characteristic
$p>0$, except that the computation is more difficult.)
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-flat-given-residue-field-extension}
Let $(A, {\mathfrak m}, k)$ be a local ring and let $k \subset k'$ be
a finite field extension. Show there exists a flat, local map of
local rings $A \to B$ such that ${\mathfrak m}_B = {\mathfrak m} B$ and
$B/{\mathfrak m} B$ is
isomorphic to $k'$ as $k$-algebra. (Hint: first do the case where
$k \subset k'$ is generated by a single element.)
\end{exercise}
\begin{remark}
\label{remark-flat-given-residue-field-extension-general}
The same result holds for arbitrary field extensions $k \subset K$.
\end{remark}
\section{The Spectrum of a ring}
\label{section-spectrum-ring}
\begin{exercise}
\label{exercise-spec-Z}
Compute $\Spec(\mathbf{Z})$ as a set and describe its topology.
\end{exercise}
\begin{exercise}
\label{exercise-basis-opens-standard}
Let $A$ be any ring. For $f\in A$ we define
$D(f):= \{\mathfrak p \subset A \mid f \not \in \mathfrak p\}$.
Prove that the open subsets $D(f)$ form a basis of the topology of
$\Spec(A)$.
\end{exercise}
\begin{exercise}
\label{exercise-radical-ideals-closed}
Prove that the map $I\mapsto V(I)$
defines a natural bijection
$$
\{\hbox{$I\subset A$ with $I = \sqrt{I}$}\}
\longrightarrow
\{\hbox{$T\subset \Spec(A)$ closed}\}
$$
\end{exercise}
\begin{definition}
\label{definition-quasi-compact}
A topological space $X$ is called {\it quasi-compact}
if for any open covering $X = \bigcup_{i\in I} U_i$ there is a finite
subset $\{i_1, \ldots, i_n\}\subset I$ such that $X = U_{i_1}\cup\ldots
U_{i_n}$.
\end{definition}
\begin{exercise}
\label{exercise-spec-quasi-compact}
Prove that $\Spec(A)$ is quasi-compact for any ring $A$.
\end{exercise}
\begin{definition}
\label{definition-Hausdorff}
A topological space $X$ is said to verify the separation axiom $T_0$
if for any pair of points $x, y\in X$, $x\not = y$ there is an open
subset of $X$ containing one but not the other.
We say that $X$ is {\it Hausdorff} if for any pair $x, y\in X$, $x\not = y$
there are disjoint open subsets $U, V$ such that $x\in U$
and $y\in V$.
\end{definition}
\begin{exercise}
\label{exercise-not-hausdorff}
Show that $\Spec(A)$ is {\bf not} Hausdorff in general.
Prove that $\Spec(A)$ is $T_0$. Give an example of a topological
space $X$ that is not $T_0$.
\end{exercise}
\begin{remark}
\label{remark-not-hausdorff}
Usually the word compact is reserved for quasi-compact and
Hausdorff spaces.
\end{remark}
\begin{definition}
\label{definition-irreducible}
A topological space $X$ is called {\it irreducible} if $X$ is not empty
and if $X = Z_1\cup Z_2$ with $Z_1, Z_2\subset X$ closed, then either
$Z_1 = X$ or $Z_2 = X$. A subset $T\subset X$ of a topological space
is called {\it irreducible} if it is an irreducible
topological space with the topology induced from $X$.
This definition implies $T$ is irreducible if and only
if the closure $\bar T$ of $T$ in $X$ is irreducible.
\end{definition}
\begin{exercise}
\label{exercise-irreducible-spec}
Prove that $\Spec(A)$ is irreducible if and only if
$Nil(A)$ is a prime ideal and that in this case it is the unique
minimal prime ideal of $A$.
\end{exercise}
\begin{exercise}
\label{exercise-irreducible-prime}
Prove that a closed subset $T\subset \Spec(A)$
is irreducible if and only if it is of the form $T = V({\mathfrak p})$ for
some prime ideal ${\mathfrak p}\subset A$.
\end{exercise}
\begin{definition}
\label{definition-generic-point}
A point $x$ of an irreducible topological space $X$ is called
a {\it generic point} of $X$ if $X$ is equal to the closure of
the subset $\{x\}$.
\end{definition}
\begin{exercise}
\label{exercise-irreducible-T0-at-most-one-generic}
Show that in a $T_0$ space $X$ every irreducible closed
subset has at most one generic point.
\end{exercise}
\begin{exercise}
\label{exercise-spec-sober}
Prove that in $\Spec(A)$ every
irreducible closed subset {\it does} have a generic point.
In fact show that the map
${\mathfrak p} \mapsto \overline{\{{\mathfrak p}\}}$ is
a bijection of $\Spec(A)$ with the set of irreducible closed
subsets of $X$.
\end{exercise}
\begin{exercise}
\label{exercise-irreducible-subset-not-generic}
Give an example to show that an irreducible
subset of $\Spec(\mathbf{Z})$ does not neccesarily have a generic point.
\end{exercise}
\begin{definition}
\label{definition-Noetherian-space}
A topological space $X$ is called {\it Noetherian} if any
decreasing sequence $Z_1\supset Z_2 \supset Z_3\supset \ldots$
of closed subsets of $X$ stabilizes.
(It is called {\it Artinian} if any increasing sequence of closed
subsets stabilizes.)
\end{definition}
\begin{exercise}
\label{exercise-Noetherian-spec}
Show that if the ring $A$ is Noetherian then
the topological space $\Spec(A)$ is Noetherian. Give an
example to show that the converse is false. (The same for
Artinian if you like.)
\end{exercise}
\begin{definition}
\label{definition-irreducible-component}
A maximal irreducible subset $T\subset X$ is called an
{\it irreducible component} of the space $X$. Such an irreducible
component of $X$ is automatically a closed subset of $X$.
\end{definition}
\begin{exercise}
\label{exercise-irreducible-in-irreducible}
Prove that any irreducible subset
of $X$ is contained in an irreducible component of $X$.
\end{exercise}
\begin{exercise}
\label{exercise-Noetherian-finite-nr-irreducible}
Prove that a Noetherian topological space $X$
has only finitely many irreducible components, say $X_1, \ldots, X_n$,
and that $X = X_1\cup X_2\cup\ldots\cup X_n$. (Note that
any $X$ is always the union of its irreducible components, but that
if $X = {\mathbf R}$ with its usual topology for instance then the irreducible
components of $X$ are the one point subsets. This is not
terribly interesting.)
\end{exercise}
\begin{exercise}
\label{exercise-irreducible-components-minimal-primes}
Show that irreducible components of $\Spec(A)$
correspond to minimal primes of $A$.
\end{exercise}
\begin{definition}
\label{definition-closed}
A point $x\in X$ is called {\it closed} if $\overline{\{x\}} = \{ x\}$.
Let $x, y$ be points of $X$. We say that $x$ is a {\it specialization}
of $y$, or that $y$ is a {\it generalization} of $x$ if
$x\in \overline{\{y\}}$.
\end{definition}
\begin{exercise}
\label{exercise-closed-maximal}
Show that closed points of $\Spec(A)$
correspond to maximal ideals of $A$.
\end{exercise}
\begin{exercise}
\label{exercise-generalization}
Show that ${\mathfrak p}$ is a generalization of ${\mathfrak q}$
in $\Spec(A)$ if and only if ${\mathfrak p}\subset {\mathfrak q}$.
Characterize closed points,
maximal ideals, generic points and minimal prime ideals in terms of
generalization and specialization. (Here we use the terminology that a point
of a possibly reducible topological space $X$ is called a generic point
if it is a generic points of one of the irreducible components of $X$.)
\end{exercise}
\begin{exercise}
\label{exercise-disjoint-closed-spec}
Let $I$ and $J$ be ideals of $A$.
What is the condition for $V(I)$ and $V(J)$ to be disjoint?
\end{exercise}
\begin{definition}
\label{definition-connected-component}
A topological space $X$ is called {\it connected} if it is nonempty and not the
union of two nonempty disjoint open subsets. A {\it connected component}
of $X$ is a maximal connected subset. Any point of $X$ is contained
in a connected component of $X$ and any connected component of $X$ is
closed in $X$. (But in general a connected component need not be open in $X$.)
\end{definition}
\begin{exercise}
\label{exercise-disconnected-spec}
Let $A$ be a nonzero ring.
Show that $\Spec(A)$ is disconnected
iff $A\cong B \times C$ for certain nonzero rings $B, C$.
\end{exercise}
\begin{exercise}
\label{exercise-connected-component-stable-generalization}
Let $T$ be a connected component
of $\Spec(A)$. Prove that $T$ is stable under generalization.
Prove that $T$ is an open subset of $\Spec(A)$ if $A$ is Noetherian.
(Remark: This is wrong when $A$ is an infinite product of copies of
${\mathbf F}_2$ for example. The spectrum of this ring consists of infinitely
many closed points.)
\end{exercise}
\begin{exercise}
\label{exercise-primes-kx}
Compute $\Spec(k[x])$, i.e., describe
the prime ideals in this ring, describe the possible specializations, and
describe the topology. (Work this out when $k$ is algebraically closed but
also when $k$ is not.)
\end{exercise}
\begin{exercise}
\label{exercise-primes-kxy}
Compute $\Spec(k[x, y])$, where $k$ is algebraically
closed.
[Hint: use the morphism
$\varphi: \Spec(k[x, y]) \to \Spec(k[x])$; if
$\varphi({\mathfrak p}) = (0)$ then localize with respect to
$S = \{f\in k[x] \mid f \not = 0\}$ and use result of lecture
on localization and $\Spec$.]
(Why do you think algebraic geometers call this affine 2-space?)
\end{exercise}
\begin{exercise}
\label{exercise-primes-Zy}
Compute $\Spec(\mathbf{Z}[y])$.
[Hint: as above.] (Affine 1-space over $\mathbf{Z}$.)
\end{exercise}
\section{Localization}
\label{section-localization}
\begin{exercise}
\label{exercise-submodule-localization}
Let $A$ be a ring. Let $S \subset A$ be a multiplicative subset.
Let $M$ be an $A$-module. Let $N \subset S^{-1}M$ be an $S^{-1}A$-submodule.
Show that there exists an $A$-submodule $N' \subset M$ such that
$N = S^{-1}N'$. (This useful result applies in particular to ideals
of $S^{-1}A$.)
\end{exercise}
\begin{exercise}
\label{exercise-localize-zero}
Let $A$ be a ring. Let $M$ be an $A$-module. Let $m \in M$.
\begin{enumerate}
\item Show that $I = \{a \in A \mid am = 0\}$ is an ideal of $A$.
\item For a prime $\mathfrak p$ of $A$ show that the image of $m$
in $M_\mathfrak p$ is zero if and only if $I \not \subset \mathfrak p$.
\item Show that $m$ is zero if and only if the image of $m$ is zero
in $M_\mathfrak p$ for all primes $\mathfrak p$ of $A$.
\item Show that $m$ is zero if and only if the image of $m$ is zero
in $M_\mathfrak m$ for all maximal ideals $\mathfrak m$ of $A$.
\item Show that $M = 0$ if and only if $M_{\mathfrak m}$ is zero
for all maximal ideals $\mathfrak m$.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-localization-is-field}
Find a pair $(A, f)$ where $A$ is a domain with three or more pairwise
distinct primes and $f \in A$ is an element such that the principal
localization $A_f = \{1, f, f^2, \ldots \}^{-1}A$ is a field.
\end{exercise}
\begin{exercise}
\label{exercise-localize-finite-module-zero}
Let $A$ be a ring. Let $M$ be a finite $A$-module. Let $S \subset A$
be a multiplicative set. Assume that $S^{-1}M = 0$. Show that there
exists an $f \in S$ such that the principal localization
$M_f = \{1, f, f^2, \ldots \}^{-1}M$ is zero.
\end{exercise}
\begin{exercise}
\label{exercise-localization-is-quotient}
Give an example of a triple $(A, I, S)$ where $A$ is a ring,
$0 \not = I \not = A$ is a proper nonzero ideal, and $S \subset A$
is a multiplicative subset
such that $A/I \cong S^{-1}A$ as $A$-algebras.
\end{exercise}
\section{Nakayama's Lemma}
\label{section-nakayama}
\begin{exercise}
\label{exercise-nakayama}
Let $A$ be a ring.
Let $I$ be an ideal of $A$.
Let $M$ be an $A$-module.
Let $x_1, \ldots, x_n \in M$.
Assume that
\begin{enumerate}
\item $M/IM$ is generated by $x_1, \ldots, x_n$,
\item $M$ is a finite $A$-module,
\item $I$ is contained in every maximal ideal of $A$.
\end{enumerate}
Show that $x_1, \ldots, x_n$ generate $M$. (Suggested solution:
Reduce to a localization at a maximal ideal of $A$ using
Exercise \ref{exercise-localize-zero} and exactness of localization.
Then reduce to the statement of Nakayama's lemma in the lectures
by looking at the quotient of $M$ by the submodule generated by
$x_1, \ldots, x_n$.)
\end{exercise}
\section{Length}
\label{section-length}
\begin{definition}
\label{definition-length}
Let $A$ be a ring. Let $M$ be an $A$-module. The
{\it length} of $M$ as an $R$-module is
$$
\text{length}_A(M)
=
\sup
\{
n
\mid
\exists\ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M,
\text{ }M_i \not= M_{i + 1}
\}.
$$
In other words, the supremum of the lengths of chains of submodules.
\end{definition}
\begin{exercise}
\label{exercise-length-is-one}
Show that a module $M$ over a ring $A$ has length $1$ if and only if
it is isomorphic to $A/\mathfrak m$ for some maximal ideal $\mathfrak m$
in $A$.
\end{exercise}
\begin{exercise}
\label{exercise-length-easy}
Compute the length of the following modules over the following rings.
Briefly(!) explain your answer. (Please feel free to use additivity of
the length function in short exact sequences, see
Algebra, Lemma \ref{algebra-lemma-length-additive}).
\begin{enumerate}
\item The length of $\mathbf{Z}/120\mathbf{Z}$ over $\mathbf{Z}$.
\item The length of $\mathbf{C}[x]/(x^{100} + x + 1)$ over $\mathbf{C}[x]$.
\item The length of $\mathbf{R}[x]/(x^4 + 2x^2 + 1)$ over $\mathbf{R}[x]$.
\end{enumerate}
\end{exercise}
\begin{exercise}
\label{exercise-compute-length}
Let $A = k[x, y]_{(x, y)}$ be the local ring of the affine plane at
the origin. Make any assumption you like about the field $k$. Suppose
that $f = x^3 + x^2y^2 + y^{100}$ and $g = y^3 - x^{999}$. What is the length
of $A/(f, g)$ as an $A$-module? (Possible way to proceed: think about the
ideal that $f$ and $g$ generate in quotients of the form $A/{\mathfrak m}_A^n=
k[x, y]/(x, y)^n$ for varying $n$. Try to find $n$ such that
$A/(f, g)+{\mathfrak m}_A^n \cong A/(f, g)+{\mathfrak m}_A^{n + 1}$
and use NAK.)
\end{exercise}
\section{Singularities}
\label{section-singularities}
\begin{exercise}
\label{exercise-singularities}
Let $k$ be any field. Suppose that $A = k[[x, y]]/(f)$ and
$B = k[[u, v]]/(g)$, where $f = xy$ and $g = uv + \delta$ with
$\delta \in (u, v)^3$. Show that $A$ and $B$ are isomorphic rings.
\end{exercise}
\begin{remark}
\label{remark-singularities}
A singularity on a curve over a field $k$ is called an
ordinary double point if the complete local ring of the curve at the
point is of the form $k'[[x, y]]/(f)$, where (a) $k'$ is a finite separable
extension of $k$, (b) the initial term of $f$ has degree two, i.e., it
looks like $q = ax^2 + bxy + cy^2$ for some $a, b, c\in k'$ not all zero, and
(c) $q$ is a nondegenerate quadratic form over $k'$ (in char 2 this means that
$b$ is not zero). In general there is one isomorphism class of such rings for
each isomorphism class of pairs $(k', q)$.
\end{remark}
\section{Hilbert Nullstellensatz}
\label{section-Hilbert-Nullstellensatz}
\begin{exercise}
\label{exercise-uncountable}
{\it A silly argument using the complex numbers!}
Let ${\mathbf C}$ be the complex number field. Let $V$ be a vector
space over ${\mathbf C}$. The spectrum of a linear operator
$T : V \to V$ is the set of complex numbers $\lambda \in {\mathbf C}$
such that the operator $T - \lambda \text{id}_V$ is not invertible.
\begin{enumerate}
\item Show that ${\mathbf C}(X) = f.f.({\mathbf C}[X])$
has uncountable dimension over ${\mathbf C}$.
\item Show that any linear operator on $V$ has a
nonempty spectrum if the dimension of $V$ is finite or
countable.
\item Show that if a finitely generated ${\mathbf C}$-algebra
$R$ is a field, then the map ${\mathbf C}\to R$ is an isomorphism.
\item Show that any maximal ideal ${\mathfrak m}$ of
${\mathbf C}[x_1, \ldots, x_n]$ is of the form
$(x_1-\alpha_1, \ldots, x_n-\alpha_n)$ for some $\alpha_i \in {\mathbf C}$.
\end{enumerate}
\end{exercise}
\begin{remark}
\label{remark-HNSS}
Let $k$ be a field. Then for every integer $n\in {\mathbf N}$ and
every maximal ideal ${\mathfrak m} \subset k[x_1, \ldots, x_n]$
the quotient $k[x_1, \ldots, x_n]/{\mathfrak m}$ is a finite field
extension of $k$. This will be shown later in the course. Of course
(please check this) it implies a similar statement for maximal ideals
of finitely generated $k$-algebras. The exercise above proves
it in the case $k = {\mathbf C}$.
\end{remark}
\begin{exercise}
\label{exercise-Hilbert-Nullstellensatz}
Let $k$ be a field. Please use Remark \ref{remark-HNSS}.
\begin{enumerate}
\item Let $R$ be a $k$-algebra. Suppose that $\dim_k R < \infty$
and that $R$ is a domain. Show that $R$ is a field.
\item Suppose that $R$ is a finitely generated $k$-algebra, and
$f\in R$ not nilpotent. Show that there exists a maximal ideal
${\mathfrak m} \subset R$ with $f\not\in {\mathfrak m}$.
\item Show by an example that this statement fails when $R$
is not of finite type over a field.
\item Show that any radical ideal $I \subset {\mathbf C}[x_1, \ldots, x_n]$
is the intersection of the maximal ideals containing it.
\end{enumerate}
\end{exercise}
\begin{remark}
\label{remark-Hilbert-Nullstellensatz}
This is the Hilbert Nullstellensatz. Namely it says
that the closed subsets of $\Spec(k[x_1, \ldots, x_n])$
(which correspond to radical ideals by a previous exercise)
are determined by the closed points contained in them.
\end{remark}
\begin{exercise}
\label{exercise-product-matrices-ring}
Let $A =
{\mathbf C}[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}]$.
Let $I$ be the ideal of $A$ generated by the entries of the
matrix $XY$, with
$$
X = \left(
\begin{matrix}
x_{11} & x_{12}\\
x_{21} & x_{22}
\end{matrix}
\right)
\quad \hbox{and} \quad
Y = \left(
\begin{matrix}
y_{11} & y_{12}\\
y_{21} & y_{22}
\end{matrix}
\right).
$$
Find the irreducible components of the closed subset $V(I)$ of
$\Spec(A)$.
(I mean describe them and give equations for each of them. You do not have
to prove that the equations you write down define prime ideals.) Hints:
\begin{enumerate}
\item You may use the Hilbert Nullstellensatz, and it suffices to find
irreducible locally closed subsets which cover the set of closed points of
$V(I)$.
\item There are two easy components.
\item An image of an irreducible set under a continuous map is
irreducible.
\end{enumerate}
\end{exercise}
\section{Dimension}
\label{section-dimension}
\begin{exercise}
\label{exercise-dimension-bigger-one-finite-nr-primes}
Construct a ring $A$ with finitely many prime ideals having dimension $> 1$.
\end{exercise}
\begin{exercise}
\label{exercise-hypersurface-in-A2-dimension-one}
Let $f \in \mathbf{C}[x, y]$ be a nonconstant polynomial.
Show that $\mathbf{C}[x, y]/(f)$ has dimension $1$.
\end{exercise}
\begin{exercise}
\label{exercise-dimension-polynomial-ring}
Let $(R, \mathfrak m)$ be a Noetherian local ring.
Let $n \geq 1$. Let $\mathfrak m' = (\mathfrak m, x_1, \ldots, x_n)$
in the polynomial ring $R[x_1, \ldots, x_n]$.
Show that
$$
\dim(R[x_1, \ldots, x_n]_{\mathfrak m'}) = \dim(R) + n.
$$
\end{exercise}
\section{Catenary rings}
\label{section-catenary}
\begin{definition}
\label{definition-catenary}
A Noetherian ring $A$ is said to be {\it catenary}
if for any triple of prime ideals
${\mathfrak p}_1 \subset {\mathfrak p}_2 \subset {\mathfrak p}_3$
we have
$$
ht({\mathfrak p}_3 / {\mathfrak p}_1) = ht({\mathfrak p}_3/{\mathfrak p}_2) +
ht({\mathfrak p}_2/{\mathfrak p}_1).
$$
Here $ht(\mathfrak p/\mathfrak q)$ means the height of
$\mathfrak p/\mathfrak q$ in the ring $A/\mathfrak q$.
\end{definition}
\begin{exercise}
\label{exercise-Noetherian-local-domain-dim-2-catenary}
Show that a Noetherian local domain of dimension $2$ is catenary.
\end{exercise}
\begin{exercise}
\label{exercise-finite-type-over-field-catenary}
Let $k$ be a field.
Show that a finite type $k$-algebra is catenary.
\end{exercise}
\section{Fraction fields}
\label{section-fraction-fields}
\begin{exercise}
\label{exercise-find-fraction-field}
Consider the domain
$$
{\mathbf Q}[r, s, t]/(s^2-(r-1)(r-2)(r-3), t^2-(r + 1)(r + 2)(r + 3)).
$$
Find a domain of the form ${\mathbf Q}[x, y]/(f)$ with isomorphic
field of fractions.
\end{exercise}
\section{Transcendence degree}
\label{section-transcendence}
\begin{exercise}
\label{exercise-algebraic-extension}